package 002；

import java.util.Scanner； //用于從指定的字符串掃描數據

public class A{

? ? ? ? ?public static void main(String[]args){

? ? ? ? ?int[] arr1={-2574,-4010,8243,771,2447,-5197,2556,8044,3314,3617,6065,-2817,3131,6318,2186,-113,629,2582,-37,-1520,164,2055,-5936,5912,1717,5988,4781,5757,892,-4394,8034,2213,-1080,-2080,5364,106,2657,566,3940,-5116,4583,1806,6555,2621,7197,528,1626,18,1049,6243,3198,4397,-1325,9087,936,-6291,662,-178,135,-3473,-2385,-165,1713,-7949,-4234,1138,2212,104,6968,-3632,3801,1137,-1296,-1215,4272,6223,-5922,-7723,7044,-2938,-8180,1356,1159,-4022,-3713,1158,-8715,-4081,-2541,-2555,-2284,461,940,6604,-3631,3802,-2037,-4354,-1213,767};//我認為把數據復制來最直觀

? ? ? ? int sum=0；

? ? ? ? for(int i=1;i<=100;i++)

? ? ? ? {

? ? ? ? ?sum+=Ai;

? ? ? ? ?}

? ? ? ? ?System.out.println("start");

? ? ? ? ?int start=nextInt();

? ? ? ? ?System.out.println("end");

? ? ? ? ?int end=nextInt;

? ? ? ? ?System.out.println(start=start+);

? ? ? ? ?System.out.println(end=end+);

? ? ? ? ?for(int i=start;i<=end;i++);/*這邊用start/end做下標(start/end是自己定義的，可以根據start/end來任取數組的范圍，像是startA20,endA66，即可實現A20到A66的求和運算*/

? ? ? ? ?sum+=Ai;

? ? ? ? ?System.out.println(sum);

}

package 002;

public class B {

public static void main(String[]args){

?int[][]a={{-2574,-4010,8243,771,2447,-5197,2556,8044,3314,3617,6065,-2817,3131,6318,2186,-113,629,-2582,-37,-1520,164,2055,-5936,5912,1717,5988,4781,5757,892,-4394,8034,2213,-1080,-2080,5364,106,2657,566,3940,-5116,4583,1806,6555,2621,-7197,528,1626,18,1049,6243,3198,4397,-1325,9087,936,-6291,662,-178,135,-3473,-2385,-165,1713,-7949,-4234,1138,2212,104,6968,-3632,3801,1137,-1296,-1215,4272,6223,-5922,-7723,7044,-2938,-8180,1356,1159,-4022,-3713,1158,-8715,-4081,-2541,-2555,-2284,461,940,6604,-3631,3802,-2037,-4354,-1213,767},{2740,-4182,-5632,-2966,-1953,567,-8570,1046,2211,1572,-2503,-1899,3183,-6187,3330,3492,-464,-2104,316,8136,470,50,466,-1424,5809,2131,6418,-3018,6002,-8379,1433,1144,2124,1624,-602,-5518,5872,870,-5175,-3961,-427,-6284,2850,481,6175,141,-766,-1897,-748,-4248,366,4823,3003,1778,3256,2182,2253,5076,5540,-2650,2451,-1875,5482,-6881,-329,-969,-8032,-2093,612,1524,-5492,5758,-7401,-5039,3241,6338,3581,4321,-1072,4942,2131,210,-7045,-7514,7450,-1142,-2666,-4485,-639,2121,-5298,-3805,-1686,-2520,-1680,2321,-4617,-1961,2076,7309}};;//定義的一個二維數組是參考javaEE基礎教程這本書上的一個二維數組的例子

a=new int[100][];//初始化a是一個長度為100的數組，a數組的數組元素又是引用類

? int sum=0;

for(int i=0;i<a.length;i++)

? ? ?for(int j=0;j<a.length;j++)

? ?sum+=a[i][j]

{

? ? ?System.out.println(a[i][j]);

}

a[0]=new int[2];//初始化a數組的第一個元素

a[0][1]=6;//訪問a數組的第一個元素所指數組的第二個數組

? ?for(int i=0;i<a[0].length;i++)//a數組的第一個元素是一個一維數組，遍歷這個一維數組

{

? ? ? ? System.out.println(sum);

}

package 002;

public static void main(String[]args)
{
int sum=0;
int[][][] a =?
{
{{-2574,-4010,8243,771,2447,-5197,2556,8044,3314,3617,6065,-2817,3131,6318,2186,-113,629,-2582,-37,-1520,164,2055,-5936,5912,1717,5988,4781,5757,892,-4394,8034,2213,-1080,-2080,5364,106,2657,566,3940,-5116,4583,1806,6555,2621,-7197,528,1626,18,1049,6243,3198,4397,-1325,9087,936,-6291,662,-178,135,-3473,-2385,-165,1713,-7949,-4234,1138,2212,104,6968,-3632,3801,1137,-1296,-1215,4272,6223,-5922,-7723,7044,-2938,-8180,1356,1159,-4022,-3713,1158,-8715,-4081,-2541,-2555,-2284,461,940,6604,-3631,3802,-2037,-4354,-1213,767}},
{{2740,-4182,-5632,-2966,-1953,567,-8570,1046,2211,1572,-2503,-1899,3183,-6187,3330,3492,-464,-2104,316,8136,470,50,466,-1424,5809,2131,6418,-3018,6002,-8379,1433,1144,2124,1624,-602,-5518,5872,870,-5175,-3961,-427,-6284,2850,481,6175,141,-766,-1897,-748,-4248,366,4823,3003,1778,3256,2182,2253,5076,5540,-2650,2451,-1875,5482,-6881,-329,-969,-8032,-2093,612,1524,-5492,5758,-7401,-5039,3241,6338,3581,4321,-1072,4942,2131,210,-7045,-7514,7450,-1142,-2666,-4485,-639,2121,-5298,-3805,-1686,-2520,-1680,2321,-4617,-1961,2076,7309}},
{{302,-7944,-6551,6648,-884,-6332,1188,-248,3902,-2171,-3782,-3510,6203,-636,-4605,4497,3187,-6278,1198,-7963,7647,-3022,-8267,7995,-1998,-4171,730,-8117,-1706,7601,-2454,-6161,-1810,243,-2841,-5377,2985,2328,177,5622,3596,3057,2830,-1151,6566,2254,6972,-3060,4063,367,84,3156,3880,-1530,5272,1385,-565,-1736,1267,-3863,-229,2072,2423,-7449,9555,2844,-1278,-3613,-919,4153,-4232,4032,-5732,199,4792,2498,-3344,-4193,2650,-579,3416,-3575,1822,-6936,-2661,-1873,414,2314,6844,7613,-7844,6680,5974,2208,-3597,6302,4396,-141,-1864,1939}}
};//三維數組的定義是同二維數組
int a=new int[100][][];//初始化a是一個長度為100的數組，a數組的元素還是引用類
int sum=0;

for(int i=0;i<a.length;i++)

? ? ?for(int j=o;j<a.length;j++)

? ? ? ? ?for(int q=o;q<a.length;q++)
? ?sum+=a[i][j][q];

a[0]=new int[2];//初始化a數組的第一個元素

a[0][1]=6;//訪問a數組的第一個元素所指數組的第二個數組

? ?for(int i=0;i<a[0].length;i++)

{

? ? ? ? System.out.println(sum);

? System.out.println(sum)；

}

心得體會和小結：

一看到題目的時候覺得很簡單，不就是分別實現一百個數，兩百個數，三百個數相加嘛，后來問了幾個同學，他們是實現幾百個數相加，但是也沒簡單。老師給的Excle表中有幾千個數，我們要實現任意范圍內的幾百個數相加。于是我就犯愁了，頓時沒了頭緒。對于在任意范圍內的幾百個數相加，那在編寫代碼的時候就該給數加上標簽，但是有那么多的數，總不能每個加上標簽吧，幾十個幾百個數也許還是依次加上標簽，但是像這樣幾千幾萬那么多的數怎么加啊，而且相加的數還不是只有一維的，有二維和三維的。不知道怎么去實現，于是我問了林杰同學，他說我可以在for循環里做文章，可以任意將i的初始值和最終的值替換，然后試著用start和end替換掉for循環里的1和100.我把第一個一維數組做完之后，發現這個方法還是挺好用的，但是在后面的二維和三維的數組上我不想我把這樣用，于是我吧上學期的javaEE的書翻出來把數組的那張看了一遍發現在處理二維數組的時候，可以將二維數組當成一維數組來遍歷二位數組里的每個元素。所以我在處理二維數組上的任意兩百個數相加的時候，參考了書上的一個例題，但是我有修改了一下，很不巧的是我的Tomcat和my eclipse壞掉了，測試不了，卸了重裝好幾次也不行。但是我知道我的代碼應該是有些問題的，基礎不是很好，編的不完美。但是我在完成一維數組任意范圍求和之后再繼續二維和三維數組，發現通過一位數組的代碼類比出來的二維和三維數組的代碼還是相對簡單的。

? ?還有就是每次的軟件工程的工程，都會迫使我去看之前沒學好的內容，溫故而知新，收獲頗多！

?

轉載于:https://www.cnblogs.com/lazygirl/p/4829427.html

總結

以上是生活随笔為你收集整理的个人项目实践的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。